2
$\begingroup$

Let $R,z > 0$ be positive real constants, and consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $$ f(v) = \frac{1}{\sqrt{(R+v)^2+z^2}}\ \mathrm{K}\!\left( \frac{4 R v}{(R+v)^2+z^2} \right)$$ where $\mathrm{K}(m)$ is the complete elliptic integral of the first kind, defined (using Wolfram Mathematica's convention) by $$ \mathrm{K}(m) = \int_0^{\pi/2} \frac{\mathrm{d} \theta}{\sqrt{1-m\sin^2 \theta}}. $$ Is $f$ an even function? Numerical tests suggest that it should be, but I can't find the right series of manipulations to demonstrate this analytically.

$\endgroup$
  • 1
    $\begingroup$ Okay, backtracking somewhat, one obvious thing to do is divide everything by $R^2$, and then you can relabel $v/R \mapsto v$, $z/R \mapsto z$ to get rid of $R$. Not much progress, but at least that leaves you fewer variables! $\endgroup$ – Chappers Apr 29 '15 at 1:00
  • $\begingroup$ Right, sorted it out (see below). $\endgroup$ – Chappers Apr 29 '15 at 1:24
3
$\begingroup$

Okay, cracked it. It's annoyingly simple in the end: $$ K(m) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m\sin^2{\theta}}} \\ = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m(1-\cos^2{\theta})}} \\ = \int_0^{\pi/2} \frac{d\theta}{\sqrt{(1-m)+m\cos^2{\theta}}} \\ = \frac{1}{\sqrt{1-m}}\int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac{m}{m-1}\cos^2{\theta}}} $$ and changing variables you conclude that $$ K(m) = \frac{1}{\sqrt{1-m}} K\left( \frac{m}{m-1} \right), $$ with caveats about square roots. Turns out you're fine, since $$ 1-m = 1- \frac{4v}{(1+v)^2+z^2} = \frac{(1+v)^2+z^2-4v}{(1+v)^2+z^2} = \frac{(1-v)^2+z^2}{(1+v)^2+z^2}, $$ clearly positive. And then you also have $$ \frac{m}{m-1} = -\frac{4v}{(1+v)^2+z^2} \frac{(1+v)^2+z^2}{(1-v)^2+z^2} = \frac{-4v}{(1-v)^2+z^2} $$

(Obviously I've set $R=1$ here, which is fine because I can just divide $v$ and $z$ by it.) Hence you end up with the evenness you asked for.

Remark: the identity I proved above is basically DLMF's 19.7.2's first equation, in different notation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That should be a $4Rv$ in the numerator of the argument of EllipticK, not $2Rv$. $\endgroup$ – David Zhang Apr 29 '15 at 0:17
  • $\begingroup$ Ah, that's more interesting... $\endgroup$ – Chappers Apr 29 '15 at 0:18
  • $\begingroup$ Fantastic! I think this is about as elegant a solution as I could have hoped for. $\endgroup$ – David Zhang Apr 29 '15 at 1:35
  • $\begingroup$ Yep, sometimes one just has to dig in and push the integrals around until they co-oporate :). Just don't ask me how to prove the AGM integral... $\endgroup$ – Chappers Apr 29 '15 at 1:39
  • $\begingroup$ For searching purposes: the relation you proved is called the imaginary modulus transformation. $\endgroup$ – J. M. isn't a mathematician Dec 30 '16 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.